The binomial series 63
+
( 1 / 2 )(− 1 / 2 )(− 3 / 2 )
3!
(x
4
) 3
+···
]
= 2
(
1 +
x
8
−
x^2
128
+
x^3
1024
−···
)
= 2 +
x
4
−
x^2
64
+
x^3
512
−···
This is valid when
∣
∣
∣
x
4
∣
∣
∣<1,
i.e. |x|< 4 or− 4 <x< 4
Problem 14. Expand
1
√
( 1 − 2 t)
in ascending
powers oftas far as the term int^3.
State the limits oftfor which the expression
is valid.
1
√
( 1 − 2 t)
=( 1 − 2 t)−
1
2
= 1 +
(
−
1
2
)
(− 2 t)+
(− 1 / 2 )(− 3 / 2 )
2!
(− 2 t)^2
+
(− 1 / 2 )(− 3 / 2 )(− 5 / 2 )
3!
(− 2 t)^3 +···,
using the expansion for( 1 +x)n
= 1 +t+
3
2
t^2 +
5
2
t^3 +···
The expression is valid when| 2 t|<1,
i.e. |t|<
1
2
or−
1
2
<t<
1
2
Problem 15. Simplify
√ (^3) ( 1 − 3 x)√( 1 +x)
(
1 +
x
2
) 3
given that powers ofxabove the first may be
neglected.
√ (^3) ( 1 − 3 x)√( 1 +x)
(
1 +
x
2
) 3
=( 1 − 3 x)
1
(^3) ( 1 +x)
1
2
(
1 +
x
2
)− 3
≈
[
1 +
(
1
3
)
(− 3 x)
][
1 +
(
1
2
)
(x)
][
1 +(− 3 )
(x
2
)]
when expanded by the binomial theorem as far as thex
term only,
=( 1 −x)
(
1 +
x
2
)(
1 −
3 x
2
)
(
1 −x+
x
2
−
3 x
2
)
when powers ofxhigher than
unity are neglected
=( 1 − 2 x)
Problem 16. Express
√
( 1 + 2 x)
√ (^3) ( 1 − 3 x)as a power
series as far as the term inx^2. State the range of
values ofxfor which the series is convergent.
√
( 1 + 2 x)
√ (^3) ( 1 − 3 x)=( 1 + 2 x)
1
(^2) ( 1 − 3 x)−
1
3
( 1 + 2 x)
1
(^2) = 1 +
(
1
2
)
( 2 x)
- ( 1 / 2 )(− 1 / 2 )
2!
( 2 x)^2 +···
= 1 +x−
x^2
2
+···which is valid for
| 2 x|< 1 ,i.e.|x|<
1
2
( 1 − 3 x)−
1
(^3) = 1 +(− 1 / 3 )(− 3 x)
(− 1 / 3 )(− 4 / 3 )
2!
(− 3 x)^2 +···
= 1 +x+ 2 x^2 +···which is valid for
| 3 x|< 1 ,i.e.|x|<
1
3
Hence
√
( 1 + 2 x)
√ (^3) ( 1 − 3 x)=( 1 + 2 x)
1
(^2) ( 1 − 3 x)−
1
3
(
1 +x−
x^2
2
+···
)
( 1 +x+ 2 x^2 +···)
= 1 +x+ 2 x^2 +x+x^2 −
x^2
2
,
neglecting terms of higher power than 2,
= 1 + 2 x+
5
2
x^2
The series is convergent if−
1
3
<x<
1
3